= Solution
Away from the shell, the reduced radial equation is free. Regularity at the origin and the definition of the S-wave <phase shift> give
$$
u_0(r)=\begin{cases}A\sin kr,&r<R,\\B\sin(kr+\delta_0),&r>R.\end{cases}
$$
The reduced radial <wave function> is continuous at $R$; otherwise its second derivative would contain a delta derivative not present in the equation. Integrating the equation across the shell gives the jump
$$
u_0'(R^+)-u_0'(R^-)=\frac{2mV_0}{\hbar^2}u_0(R).
$$
Divide by the common shell value and substitute the two free solutions. This yields
$$
\boxed{k\cot(kR+\delta_0)-k\cot(kR)=\frac{2mV_0}{\hbar^2}}.
$$
If the shell value vanishes, use the undivided matching equations or a continuous limit rather than this logarithmic-derivative division. For an explicit <phase shift of a delta-shell potential>, write $x=kR$ and $g=2mV_0R/\hbar^2$. The cotangent addition formula gives
$$
\boxed{\tan\delta_0(k)=-\frac{g\sin^2x}{x+g\sin x\cos x}}.
$$
The phase is defined modulo $\pi$; choose its branch continuously as energy or coupling varies. At zero coupling the physical <phase shift> is zero.
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